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加密过程中,将两个待加密的图像加密成两个均匀的白噪声。待加密的图像为非负分布o(x, y),首先给予待加密图像初始随机相位P1(x, y)=exp[i2π×r(x, y)],其中,r(x, y)表示0~1的随机分布数值。同时也作为加密密钥。相应的复振幅表达为:
$ {O_1}\left( {x,y} \right) = o\left( {x,y} \right){P_1}\left( {x,y} \right) $
(1) 根据作者的设计思想,令上述复振幅是由两个具有随机相位分布的复光场的傅里叶变换场的相干叠加产生,即:
$ \begin{array}{*{20}{c}} {{O_1}\left( {x,y} \right) = \mathscr{F}\left[ {A\left( {\varepsilon ,\eta } \right){P_2}\left( {\varepsilon ,\eta } \right)} \right] + }\\ {\mathscr{F}\left[ {B\left( {\varepsilon ,\eta } \right){P_2}\left( {\varepsilon ,\eta } \right)} \right]} \end{array} $
(2) 式中,$ \mathscr{F} $表示傅里叶变换,A(ε, η)和B(ε, η)是频谱面上的振幅,P2(ε, η)为相位。第2步是将频谱面上的A(ε, η)和B(ε, η)经过随机相位板P3(ε, η)=exp[i2π×r(ε, η)]进行加密,获得最终的加密图像o1′(x1, y1)和o2′(x1, y1):
$ \begin{array}{*{20}{c}} {{{o'}_1}\left( {{x_1},{y_1}} \right){R_1}\left( {{x_1},{y_1}} \right) = \mathscr{F}\left[ {{o_1}\left( {\varepsilon ,\eta } \right)} \right] = }\\ {\mathscr{F}\left[ {A\left( {\varepsilon ,\eta } \right){P_3}\left( {\varepsilon ,\eta } \right)} \right]} \end{array} $
(3) $ \begin{array}{*{20}{c}} {{{o'}_2}\left( {{x_1},{y_1}} \right){R_2}\left( {{x_1},{y_1}} \right) = \mathscr{F}\left[ {{o_2}\left( {\varepsilon ,\eta } \right)} \right] = }\\ {\mathscr{F}\left[ {B\left( {\varepsilon ,\eta } \right){P_3}\left( {\varepsilon ,\eta } \right)} \right]} \end{array} $
(4) 所以,o1′(x1, y1)和o2′(x1, y1)就是包含待加密图像信息的两个白噪声振幅,即完成了加密过程;而R1(x1, y1)和R2(x1, y1)将作为解密的密钥。整个加密过程不需要任何迭代运算。
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解密系统如图 1所示。它是一个在频谱面放置全息光栅的4f系统。将获得的两个包含带加密图像信息的白噪声o1′(x1, y1)和o2′(x1, y1)放置在4f系统的输入面Σ上,两者的间距为2b,同时在输入面上放置解密密钥,分别为R1(x1, y1)和R2(x1, y1)。
根据图 1所示,首先在输入面Σ上获得o1′(x1, y1)R1(x1, y1)和o2′(x1, y1)R2(x1, y1)。经过傅里叶变换,在频谱面Σ0上获得:
$ \begin{array}{*{20}{c}} {g\left( {\varepsilon ,\eta } \right) = \mathscr{F}\left[ {{{o'}_1}\left( {{x_1} + b,{y_1}} \right){R_1}\left( {{x_1} + b,{y_1}} \right) + } \right.}\\ {\left. {{{o'}_2}\left( {{x_1},{y_1}} \right){R_2}\left( {{x_1} + b,{y_1}} \right)} \right] = }\\ {{o_1}\left( {\varepsilon ,\eta } \right)\exp \left[ {{\rm{i}}2{\rm{ \mathsf{ π} }}b\varepsilon } \right] + {o_2}\left( {\varepsilon ,\eta } \right)\exp \left[ {{\rm{i}}2{\rm{ \mathsf{ π} }}b\varepsilon } \right]} \end{array} $
(5) 在频谱面Σ0上,放置相位密钥R3(ε, η)和一个全息光栅,该全息光栅的透过率函数为:
$ \begin{array}{*{20}{c}} {H\left( {\varepsilon ,\eta } \right) = \frac{1}{2} + \frac{1}{4}\exp \left[ {{\rm{i}}\left( {2{\rm{ \mathsf{ π} }}{\varepsilon _0}\varepsilon } \right)} \right] + }\\ {\frac{1}{4}\exp \left[ {{\rm{i}}\left( { - 2{\rm{ \mathsf{ π} }}{\varepsilon _0}\varepsilon } \right)} \right]} \end{array} $
(6) 式中,ε0为全息光栅的空间频率。在解密过程中,在频谱面上放置的解密密钥为:
$ {R_3}\left( {\varepsilon ,\eta } \right) = {P_3}^ * \left( {\varepsilon ,\eta } \right) \times {P_2}\left( {\varepsilon ,\eta } \right) $
(7) 式中,*表示共轭。在频谱面Σ1上,经过相位板R3(ε, η)和全息光栅后,为了获得解密图像,光栅的参量ε0和b满足b=λfε0,其中, λ为照射波长,f为透镜焦距。然后光场分布为:
$ \begin{array}{*{20}{c}} {{g_1}\left( {\varepsilon ,\eta } \right) = g\left( {\varepsilon ,\eta } \right){R_3}\left( {\varepsilon ,\eta } \right)H\left( {\varepsilon ,\eta } \right) = }\\ {\frac{1}{2}g\left( {\varepsilon ,\eta } \right){R_3}\left( {\varepsilon ,\eta } \right) + \frac{1}{4}\exp \left[ {{\rm{i}}\left( { - 2{\rm{ \mathsf{ π} }}{\varepsilon _0}\varepsilon } \right)} \right] \times }\\ {{o_1}\left( {\varepsilon ,\eta } \right)\exp \left[ { - {\rm{i}}2{\rm{ \mathsf{ π} }}b\varepsilon } \right] + \frac{1}{4}\exp \left[ {{\rm{i}}\left( { - 2{\rm{ \mathsf{ π} }}{\varepsilon _0}\varepsilon } \right)} \right] \times }\\ {{o_1}\left( {\varepsilon ,\eta } \right)\exp \left[ {{\rm{i}}2{\rm{ \mathsf{ π} }}b\varepsilon } \right] + \frac{1}{4}\exp \left[ {{\rm{i}}\left( {2{\rm{ \mathsf{ π} }}{\varepsilon _0}\varepsilon } \right)} \right] \times }\\ {{o_1}\left( {\varepsilon ,\eta } \right)\exp \left[ {{\rm{i}}2{\rm{ \mathsf{ π} }}b\varepsilon } \right] + \frac{1}{4}\exp \left[ {{\rm{i}}\left( { - 2{\rm{ \mathsf{ π} }}{\varepsilon _0}\varepsilon } \right)} \right] \times }\\ {{o_1}\left( {\varepsilon ,\eta } \right)\exp \left[ {{\rm{i}}2{\rm{ \mathsf{ π} }}b\varepsilon } \right]} \end{array} $
(8) 此后,(8)式的光场再经过一次傅里叶变换便获得输出,如下式所示:
$ \begin{array}{*{20}{c}} {\mathscr{F}\left[ {{g_1}\left( {\varepsilon ,\eta } \right)} \right] = \frac{1}{2}o\left( {x,y} \right){P_1}\left( {x,y} \right) + }\\ {\frac{1}{4}{{o'}_1}\left( {x - b,y} \right){R_1}\left( {x - b,y} \right) * \mathscr{F}\left[ {{R_3}\left( {\varepsilon ,\eta } \right)} \right] + }\\ {\frac{1}{4}{{o'}_2}\left( {x + b,y} \right){R_2}\left( {x + b,y} \right) * \mathscr{F}\left[ {{R_3}\left( {\varepsilon ,\eta } \right)} \right] + }\\ {\frac{1}{4}{{o'}_1}\left( {x - 2b,y} \right){R_1}\left( {x - 2b,y} \right) * \mathscr{F}\left[ {{R_3}\left( {\varepsilon ,\eta } \right)} \right] + }\\ {\frac{1}{4}{{o'}_2}\left( {x + 2b,y} \right){R_2}\left( {x + 2b,y} \right) * \mathscr{F}\left[ {{R_3}\left( {\varepsilon ,\eta } \right)} \right]} \end{array} $
(9) 式中, *表示卷积。由(9)式可见,在输出面上一共获得5项输出,分别对应于(8)式中的每一项的傅里叶变换;并且它们的位置分别是(0, 0),(b, 0),(-b, 0),(2b, 0),(-2b, 0)。其中位于中心的图像为o(x, y)×P(x, y),取振幅即为原始图像,因此完成了解密过程。
最终的加密图像为两个实值的平稳白噪声,利于保存和传输,因此该方法的加密信息不需要全息存储;解密过程对系统的附加参量敏感,这些系统参量也可作为密钥。
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采用计算机对所提出的方法进行了模拟验证和分析。待加密图像如图 2a所示。像素为256×256,图像的大小为3cm×3cm。实验中选用的激光波长为633nm,频谱面上的光栅的空间频率ε0=10/mm,透镜的焦距为0.2m。因此根据上面的分析,解密中输入平面的两幅图的距离b应为1.27mm。首先,根据第1节中描述的加密过程,获得了两个白噪声如图 2b和图 2c所示,可见已经完全隐藏了待加密图像的信息和特征。随后,利用全部正确的密钥,包括相位板R1,R2和R3,光栅的透过率函数H,波长λ,光场常数ε0和两个加密图像的距离2b,可获得的解密图像如图 2d所示,可见当作者的密钥都正确时,能够很好地还原原始图像。若采用错误的密钥,如仅有密钥R1错误而其它密钥全部正确的情况下,所获得的解密结果图 2e所示;而若仅有密钥R2错误而其它密钥全部正确的情况下,所获得的解密结果图如图 2f所示; 若仅有密钥R3错误而其它密钥全部正确的情况下,所获得的解密结果图如图 2g所示。可见在上述3种情况下的解密结果类似白噪声,完全得不到原始图像的有效信息。由此可知,相位密钥R1, R2和R3均能够分别独立地保证该加密方法的安全性。
Figure 2. a—the image to be encrypted b—the encrypted image o1 c—the encrypted image o2 d—decryption result with correct key e—decryption result with wrong key R1 f—decryption result with wrong key R2 g—decryption result with wrong key R3
为了定量比较解密图像与原始图像的相似性,作者采用解密图像和原图像均方误差Re进行评价:
$ {R_{\rm{e}}} = \frac{{\sum\limits_{n = 1}^N {\sum\limits_{m = 1}^N {{{\left| {\left| {o'\left( {m,n} \right)} \right| - \left| {o\left( {m,n} \right)} \right|} \right|}^2}} } }}{{\sum\limits_{n = 1}^N {\sum\limits_{m = 1}^N {{{\left| {o\left( {m,n} \right)} \right|}^2}} } }} $
(10) 式中, N×N为图像的像素点数,在这里是256×256,o(m, n)和o′(m, n)分别表示原图像和解密图像的振幅,将图 2a和图 2d中的振幅数据带入该公式得到相应的数值为2.21×10-1,是一个很小的值。而当密钥错误时,将图 2e、图 2f和图 2g中的振幅数据分别带入(10)式所得到的Re值分别为0.41, 0.42和0.45,这表明重构结果与原始结果无关联。
此外,该图像加密系统对参量b和f的敏感性进行了测试和计算,如图 3所示。图 3a和图 3b中分别为图像距离b和透镜焦距f与正确值有差别时,解密图像和原图像均方误差Re的变化曲线。
由图 3可见,图像距离b和透镜焦距f距正确值有微小变化时,Re迅速增大,所以这些系统参量能够增强该系统的安全性。
基于光栅滤波器的新型光学图像加密技术
New optical image encryption technology based on grating filter
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摘要: 为了提高干涉加密技术的光学处理速度和光学实现的可行性,基于干涉原理的加密技术的思想,结合光栅滤波器的图像处理方法,研究了基于光栅滤波器的4f系统双图像光学加密技术。该技术利用已成熟的基于光栅滤波的图像相加减4f系统,将两个待加密图像转化为两个实值的白噪声,理论模拟验证了该方法的可行性和有效性。结果表明,该方法简单实用,易于光学实现并具有很高的安全性。Abstract: In order to improve optical processing speed and the feasibility of optical realization in interference encryption technology, based on the idea of interference encryption principle and image processing method of grating filter, 4f system double image optical encryption technology based on grating filter was designed. The mature image additiion and subtraction 4f system based on grating filtering was used to transform two encrypting images into two real values of white noise. The feasibility and effectiveness of the method were verified by theoretical simulation. The results show that the method is simple, practical, safe and easy to implement.
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Key words:
- information optics /
- optical encryption /
- image encryption /
- grating filter /
- interference
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